# Rings with Unity

• May 2nd 2009, 11:38 AM
tfhawk
Rings with Unity
I know that unity for a ring is just the multiplicative inverse for a ring, and that not every ring has a multiplicative inverse, so therefore not every ring has unity. I need to prove why a ring has at most one unity. I'm just not sure how to construct a well written proof for this. Any help would be greatly appreciated.
• May 2nd 2009, 12:28 PM
TheEmptySet
Quote:

Originally Posted by tfhawk
I know that unity for a ring is just the multiplicative inverse for a ring, and that not every ring has a multiplicative inverse, so therefore not every ring has unity. I need to prove why a ring has at most one unity. I'm just not sure how to construct a well written proof for this. Any help would be greatly appreciated.

I'm not quite what you mean by multiplicative inverse for a ring?

Unity is the multiplicative Identity

PlanetMath: unity

The idea of the proof is

what if there are to elements of the Ring that are the multiplicative Identity call them $e_1,e_2$

but then

$e_1 \cdot e_2 = e_2$ becuase $e_1$ is an identity, and

$e_1 \cdot e_2 = e_1$ becuase $e_2$

This implies that

$e_1=e_1 \cdot e_2=e_2 \implies e_1=e_2$

So the identity is unique.
• May 2nd 2009, 01:03 PM
Gamma
I think you are confusing the term "unit" and "unity" which is understandable. A unit is an element in a ring that has an inverse in the ring (multiplicative).
unity is the identity element 1, such that 1r=r for all r in your ring. It is indeed uniques as $\emptyset$ pointed out.

It should also probably be noted that rings can certainly have unity without every element having an inverse. For instance $\mathbb{Z}$ is a ring with unity 1, but only 1 and -1 have inverses.

This is also a typical example to show that not every element in a ring is either a unit (invertible) or a zero divisor. This is an integral domain and has no zero divisors and only 2 units.